Add solution #542

Author: JoshCrozier

README.md

@@ -117,6 +117,7 @@
 500|[Keyboard Row](./0500-keyboard-row.js)|Easy|
 506|[Relative Ranks](./0506-relative-ranks.js)|Easy|
 541|[Reverse String II](./0541-reverse-string-ii.js)|Easy|
+542|[01 Matrix](./0542-01-matrix.js)|Medium|
 551|[Student Attendance Record I](./0551-student-attendance-record-i.js)|Easy|
 557|[Reverse Words in a String III](./0557-reverse-words-in-a-string-iii.js)|Easy|
 565|[Array Nesting](./0565-array-nesting.js)|Medium|

solutions/0542-01-matrix.js

@@ -0,0 +1,45 @@
+/**
+ * 542. 01 Matrix
+ * https://leetcode.com/problems/01-matrix/
+ * Difficulty: Medium
+ *
+ * Given an m x n binary matrix mat, return the distance of the nearest 0 for each cell.
+ *
+ * The distance between two adjacent cells is 1.
+ */
+
+/**
+ * @param {number[][]} matrix
+ * @return {number[][]}
+ */
+var updateMatrix = function(matrix) {
+  const queue = [];
+
+  for (let i = 0; i < matrix.length; i++) {
+    for (let j = 0; j < matrix[0].length; j++) {
+      if (matrix[i][j] === 0) {
+        queue.push([i, j, 0]);
+      } else {
+        matrix[i][j] = Infinity;
+      }
+    }
+  }
+
+
+  while (queue.length) {
+    const [i, j, k] = queue.shift();
+    if (matrix[i][j] > k) {
+      matrix[i][j] = k;
+    }
+    [[-1, 0], [1, 0], [0, -1], [0, 1]].forEach(p => {
+      const [x, y, z] = [i + p[0], j + p[1], k + 1];
+      if (x > -1 && x < matrix.length && y > -1 && y < matrix[0].length) {
+        if (matrix[x][y] === Infinity) {
+          queue.push([x, y, z]);
+        }
+      }
+    });
+  }
+
+  return matrix;
+};